LO 01.2.01
. '<>llttC the gene that controls the
· · t 1 ·1vc been nbl c t0 I s
.
insert
the gene into another type of
Using special enzymes, sctcntJs s "
.
fp
. A fl·om one type of bnctcnn IIIH1 ,
productiOn o rotem
t
(l
. A These b·H.:tcria then produce Protein A.
bacteria that does not nonnally produce rotclll ·
·
•
What is the name of this technique?
A- Cloning
B- Meiosis
C- Genetic modification
c
LO 81.2.01
GM crops are different from anything that has existed before because:
They are produced by techniques of cross-breeding
They are capable of drastically improving agricultural yields
They involve .transplanting genes between different organisms
LO 81.2.01
Which of the following is not an advantage of producing genetically modified food?
A- Producing greater yield within an shorter period of time.
B- Producing cereals which contain a variety of nutrients.
C- Increasing the resistance to pests and thus reducing the usc of pesticides
D- Upsetting the equilibrium in the ecosystem .
LO 81.2.01
Large quantities ofuseful products can be produced through genetic engineering involving:
A- bacteria containing recombinant plasmids
B- yeast carrying foreign genes
;· ~ · ;.
C- transgenic plants
D- all ofthe above
...
BT com is a type of genetic modified crops that is a type of com modified with Bt bacteria,
the reason behind this modification
A- To increase nutrition in com for developing countries
B- To decrease the wastes comes from the com fields
C- To get rid of European com borer which take out 5% of the field each year
D- To increase resistance to pests
BIOLOGY - Grade 11
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LO 81.2.01
Which of the followii1g is not an advantage of producing genetically
modified food'!
AProducing greater yield within a short period of time.
n- Producing cereals which contain a variety of nutrients.
Increasing the resistance to pests and thus reducing the use of
Cpesticides
U
· the
LO 81.2.02
NliTRJENT AGAR PLATES
In a transfommtion experiment, a sample
Ampicillin
No Ampicillin
of E. coli bacteria was mixed with a plasmid
containing the gene for resistance to the antibiotic
ampicillin (ampr). Plasmid was not added to
-;·Wild-lype
· E. coli
a second sample. Samples were plated on
nutrient agar plates, some of which were
II
supplemented with the antibiotic ampicillin.
The results of E. coli growth are summarized
E. coli and
below. The shaded ar~a represents
anrpr plasmld
.
extensive growth of bacteria; dots represent
IV
individual colonies of bacteria.
u
0
.
Use this information to Answer questions 8 and 9:
Plates that have only ampicillin-resistant bacteria growing include
which of the following?
AB-
CD-
I only
Ill only
IV only
I and I
LO 81.2.02
I. Plates I and III were included in the experimental design in order to
A- Demonstrate that the E. coli cultures were viable
B- Demonstrate that the plasmid can lose its ampr gene
C- Demonstrate that the plasmid is needed for E. coli growth
are the E. coli for transfonnation
LO 81.2.02
2. How can you check to see if the GFP gene was moved into the bacteria
Aa. Count how many bacteria you have on your plate
Bb. See if the bacteria changed shape
Cc. Observe the bacteria under the microscope.
Dd. Look at the bacteria under UV light.
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· that produce human ·msu1·m, scten
· f lSts insert the human gene of interest
To create bactena
into the bacterial:
A- Circular chromosome
B- Plasmid
C- Cell wall
DLO 81.2.02
Use the following information to answer the question below.
A eukaryotic gene has "sticky ends" produced by the restriction endonuclease EcoRI. The
gene is added to a mixture containing EcoRI and a bacterial plasmid that carries two genes,
which make it resistant to ampicillin and tetracycline. The plasmid has one recognition site
for EcoRI located in the tetracycline resistance gene. This mixture is incubated for several
hours and then added to bacteria growing in nutrient broth. The bacteria are allowed to
grow overnight and are streaked on a plate using a technique that produces isolated
colonies that are clones of the original. Samples of these colonies are then grown in four
different media: nutrient broth plus ampicillin, nutrient broth plus tetracycline, nutrient
broth plus ampicillin and tetracycline, and nutrient broth containing no antibiotics. The
bacteria containing the engineered plasmid would grow in
--AAmpicillin and tetracycline broth only.
B- Nutrient broth, the ampicillin broth, and the tetracycline broth.
C- Nutrient broth and the tetracycline broth only.
icillin broth and the nutrient broth.
Asexual reproduction produces offspring that each contain
AGenetic information from one parent
B- Genetic information from two parents
C- Less genetic information than either parent
A un · ue combination of
· information
LO BI.2.02
Homologus chromosomes include
AOne smaller and one bigger chromosome.
B- One chromosome from each parent.
C- One complete and one incomplete chromosome.
D- None of the above.
Which of the following is NOT true concerning mitosis?
AB-
CD-
Animal cells have centrioles while plant cells do not.
Both plant and animal cells undergo cytokinesis.
Mitosis allows growth and increase in size in both plants and animals.
Animal cells form a cell plate during c~okinesis
while plant cells do not.
.
BIOLOGY -Grade 11
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LO 81.2.02
The term thnt refers to division of the cytoplasm is:
ACytokinesis
UAnaphase
CApoptosis
kinesis
Generally, complex organisms do require more genes to control their
synthesis and organization than do primitive orgat:lisms. However, the numbers of
chromosomes vary from ants with 2, molds with··s(f4, humans with 46, potatoes with
100 and the crayfish with 200, Given this, then
A- There must be no relationship between amount of genetic information
and complexity of the organism.
B- The number of genes per chromosome may vary among organisms,
preventing a simple relationship between chromosome number and complexity.
C- Mitosis must differ from organism to organism.
Simpler
have more DNA than more
LO Bl.2.03
In a Mendelian monohybrid cross, which generation is always completely
homozygous?
AF 1 generation
BF 2 generation
CF 3 generation
DP generation
LO Bl.2.03
A cross between two true breeding lines one with .dark blue flowers and one with bright
white flowers produces F 1 offspring that are light plhe. When the F 1 generation is mated a
1 :2:1 ratio of dark blue to light blue to white flowers is observed. What genetic
phenomenon is consistent with these results?
AEpistasis
BIncomplete dominance
CCo dominance
lnbreedi
LO Bl.2.03
A gene showing co-dominance:
AHas both alleles independently expressed in the heterozygote
BHas one allele dominant to the other
CHas alleles tightly linked on the same chromosome
Has alleles expressed at the same time in development
D-
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IIIII :J fj,IC
-21- 8
i
. cros~ed WI·than individual with a
If an individual with a dominant phenotype IS
recessive phenotype, 4 of their 9 offspring show the recessive phenotype.
What is the genotype of the first parent?
AB-
CD-
AA
Aa
aa
AA or Aa
I : II' jE
:
lllej:Jfjlll .
Unattached earlobes (EE or Ee) are dominant over attached earlobes (ee). A couple both h
ave unattached earlobes. Both notice that one of their parents on both sides has attached
earlobes (ee). Therefore, they correctly assume that they are carriers for attached earlobl!s
(Ee). The couple proceed to have four children.
A
8
C
D
I
Th~y can be certain that three will be heterozygous and one homozygous rcc
esstve.
If the first three are heterozygous, the fourth must be homozygous
recessive.
All children must have unattached earlobes since both parents possess
the dominant gene for it.
..
Two heterozygous, one homozygous recessive and one homozygous domina
nt is a likely outcome, but all heterozygous, or two, three or all four
homozygous are also possible.
QUESTION 23-25
In Zea mays, the allele for coloured seed (C) is dominant over the allele for colourless
seed (c). The allele for starchy endosperm (S) is dominant over the allele for waxy
endosperm (s ). Pure breeding plants with coloured seeds and starchy endosperm were
crossed with pure breeding plants with colourless seeds and waxy endo~erm .
cState
.: ,tJc
:>c: : =
''''':''l"t:
the genotype of the F 1 individuals produced as a result of this cross:
ABC0-
'
SSCC
SsCc
Sscc
sscc
·24 •
_ _ _...._......... ... ~ - · I LO B l . 2 . 0 3 In colorless seed with waxy endosperm parent, the percentage of the gametes that
carries the allele of dominance
·
A-
12
BC0-
y.
%
zero
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LO BI.2.03
The observed percentages of phenotypes in the F2 generation are shown below.
coloured starchy 126
colourless starchy 44
coloured waxy42
colourless waxy 14.
What is the percentage of the F2?
A3:3:3:3
B9:3:3: I
Cl : l : 1: 1
D3:1
LO Bl.2.03
Human blood types are genetically detennined. The table below shows the symbols used to
represent two of the alleles for blood types and gives a description of each allele.
Symbol
I"
••
Allrle Dncrlption
produc~
antigen A on roo
bJoOO c~lls
produces antigen B on roo
blood Q?lll
Two Alleles Controlling Human Blood Type
In homozygous individuals, two lA alleles result in blood type A and two IB alleles result
in blood type B. The IA and IB alleles are co-dominant, resulting in blood type AB in
individuals heterozygous for the two alleles. A male and a female both have blood type
AB. If they have a child, what is the probability that the child will also have blood type
AB?
A1/4
B1/2
C3/4
DIll
27
I
LO BI.2.03
In the fruit fly Drosophila melanogaster, vestigial wings and hairy body are produced by
two recessive genes located on different chrom.os~mes. The normal alleles, long wings and
hairless body, are dominant. They givelOO% Fl progeny obtained from a cross between a
vestigial-winged, hairy male and a normal, homozygous female. If the Fl from this cross
are permitted to mate randomly among themselves,
-What phenotypic ratio would be expected in the F2 generation?
9 normal long wing, hairless body: 3 vestigial wing, hairless body: 3 nonnal
Along wing, hairy body:
1 vestigial wing, hairy body
3 nonnal long wings ,hairless body : 3 normal long wing , hairy body
B: I vestigial wings hairless body : 1 vestigial wings ,hairy body
1normal long wing, hairless body: I vestigial wing, hairless body
C: 1 normal long wing, hairy body: 1 vestigial wing, hairy body
D100% nonnal long wings ,hairless body
BIOLOGY - Grade 11
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LO 81.2.04
your answer to the following question on the pedigree chart below, which shows
a history of ear lobe shape
KEY
E=Allele for free ear lobes (dominant)
e=Allele for attached ear lobes (recessive)
0 =Male with free ear lobes
0 =Female with free ear lobes
I =Male with attached ear lobes
I =Female with attached ear lobes
What could the genotype of individual 1 be?.
AEE, only
BEe, only .
Cee
DEE or Ee
Study the following pedigree to Answer question 29&30
LO BI.2.0"
Deduce what is the type of inheritance of this disease
AAutosomal recessive
BAutosomal Dominance
X- linked recessive trait
CDLinked trait
LO 81.2.0-'
From the previous pedigree, the individual (13) genotype could be
AXaX3 only
BXA XA only
CCould be xAxu or XAXA
DXAX11 only
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/
'
-Jl-
LOBI.2.0 4 -
Using the pedigree below to deduce the type of inheritance in this family
A-
8CD-
-
Autosomal recessive
Sex-linked recessive
Autosomal dominant
Linked gene
:n -
LO 81.2.0-t
Scientist study the genome of bacteria in termites digestive tracts , to find answer about
breaking down of cellulose to produce ethanol , this is an example of applied genome in
field of
AAlternative energy
8Ecosystem biodiversity
CEvolution
DHuman Health
LOBI.2.0 . t -
- 3.l -
How does the X chromosome differ from the Y chromosome in humans?
A-
The Y chromosome is longer.
8-
Some genes on the X chromosome are absent from the Y chromosome.
C-
The genes are the same but some on the Y chromosome are not expressed.
D-
The X chromosome determines sex.
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Key
0
0
=affected
= unaffected
LO BI.2.04
According to the pedigree diagrammed above the original parents of this family
tree had:
A3 daughters and 2 sons
B2 daughters and 3 sons
C2 daughters and 1 son
1 dau
and 2 sons
LO 81.2.04
Which of the statements below is/are true regarding the trait of interest in the above
pedigree?
If it is autosomal recessive, all the chlidren of the affected daughter would be
Aaffected as well.
If it is X-linked recessive, all the children of the affected daughter would be af
Bfected.
It is likely autosomal dominant; the affected individuals would be heterozygo
Cus.
All of these statements are true.
DLO Bl.2.04
Identify the correct order of organization of genetic material, from largest to
smallest.
AGene, chromosome, nucleotide, genome
Chromosome, gene, genome, nucleotide
Chromosome, genome, nucleotide, gene
Genome, chromo
nucleotide
LO 81.2.04
Scientist study the genome of bacteria in termites digestive tracts , to find answer about
breaking down of cellulose to produce ethanol , this is an example of applied genome in
field of
AAlternative energy
8Ecosystem biodiversity
Evolution
CHuman
Health
DBIOLOGY - Grade 11
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./
/
During which phase of the cell cycle d
oes
DN
. .
LO Bl.2.05
A rephcanon occur?
I
ABC-
Prophase·
Metaphase
Anaphase
LO B1.2.05
The sequence of one strand of DNA is 5 TCGA TG 3 . The sequence of the complementary
strand would be.
A3' AGCTAC 5'
B5' TCGATC 3'
C5' CTAGCT 3'
D3'GCTAGC 5'
LO BI.2.05
IN a DNA sample, the percentage of thymine is 20% .Then what will be the
percentage of guanine?
A20%
B40%
C-
Starting with N
(heavy) DNA and after 2 generation in N medium . E.coli cell
will contain
14
15
15
14
14
Aa. 25% N 15 N DNA, 50% N N DNA ,and 25% N N DNA
14
14
Bb. 50% N 15 N 15 DNA and 50% N N DNA
Cc. 50% N 15 N 15 DNA and 50% N 15N 14 DNA
14
14
14
Dd. 50% N 15 N DNA and 50% N N DNA
BIOLOGY- Grade 11
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.
.
tions (42-44} on the diagram bclo"v which
atld 011 your knowledge of
Base your answer to the followmg gucs
. of a doubl e-s tranded DNA mo 1ecu 1e '
represents a portton
biology
If
I
LO 81.2.05
The base sequence of strand II is most likely:
A3' C-A-C-T-G-G 5 '
B3' G-G-T:.C-A-C5 '
CS'G-T-G-A-C-C3' .
D3' G-T-G-A-C-CS'
LO BI.2.05
Which carbon in molecule No.3 will bind to the nitrogenous base ( C):
ACarbon No:4
BCarbon No:3
CCarbon No: 5
DCarbon No: 1
I LO BI.2.05
When bonded together chemically, deoxyribose, phosphate, and an adenine
molecule make up
ADNA nucleotide
BRNA nucleotide
CDNA molecule
DAn RNA molecule
I LO BI.2.06
1. Which of the following best describes the formation of a zygote?
AB-
CD-
A sperm cell nucleus and an egg cell nucleus fuse.
A cell's DNA replication and mitosis are accelerated.
A succession of cell divisions produces a solid mass of cells.
A cel1 with 46 chromosomes divides tofom1 cells with 23
chromosomes each.
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LO 81.2.06
Base your answer
to .the. fall owmg
· question
· 011 The d1agram
.
.
below shows a process that
can occur dunng meiOSIS
The most likely result of this process is
Aa new combination of inheritable traits that can appear in the offspring
Ban inability to pass either of these chromosomes on to offspring
.
a loss of genetic information that will produce a genetic disorder in the offspn
Cng
an increase in the chromosome number of the organism in which this process
Doccurs
LO BI.2.06
3. Which diagram illustrates fertilization that would most likely lead to the development of
a abnormal human female?
A)
C)
@+~
@+~
B)
D)
@+~
@+~
a
AB-
b
C-
c
D-
d
LO 81.2.06
Hemophilia is an X-linked recessive disorder. A daughter can be afflicted with hemophilia if
she inherits
.
. .. , ·., r
Atwo normal X chromosomes; hemophilia usually arises spontaneously
Ban Xh allele from her mother only.
an Xh allele from her father and her mother
CDan Xh allele from her father only
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LO 81.2.06
4.
Wild type fmit flies have red eyes. A white-eyed female fly is crossed with a red-eyed
male fly. All of the females from the cross are red-eyed and all of the males, white-eye
d. What type of inheritance pattern is this?
AAutosomal dominant
BAutosomal recessive
CIncomplete dominance
DSex-linked on X chromosome
LO BI.2.06
If a female inherits and expresses an X-linked recessive disorder, what must be true about her
parents?
ABoth parents must have the disease
BHer must father has the disease.
CNeither parent has the disease.
DHer mother has the disease
BEST OF LUCK
BIOLOGY- Grade 11
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